SOLUTION: John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his patio (area of a
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-> SOLUTION: John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his patio (area of a
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Question 51780: John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation.
Show clearly the algebraic steps which prove your dimensions are the maximum area which can be obtained. Use the vertex form to find the maximum area. Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! 2l + 2w = 300
l = (300 - 2w) / 2
l = 150 - w
A = w*l
A = w*(150 - w)
Let's say w = 75 (square patio)
What if w = 74?
What if w = 76?
So, if I go up a little from w = 75 or down
a little from w = 75, A goes down.
w = 75 and l = 75 gives the max area
